HDOJ 1266 Reverse Number（数字反向输出题）

Problem Description

Welcome to 2006’4 computer college programming contest!

Specially, I give my best regards to all freshmen! You are the future of HDU ACM! And now, I must tell you that ACM problems are always not so easy, but, except this one… Ha-Ha!

Give you an integer; your task is to output its reverse number. Here, reverse number is defined as follows:

1. The reverse number of a positive integer ending without 0 is general reverse, for example, reverse (12) = 21;

2. The reverse number of a negative integer is negative, for example, reverse (-12) = -21;

3. The reverse number of an integer ending with 0 is described as example, reverse (1200) = 2100.

Input

Input file contains multiple test cases. There is a positive integer n (n<100) in the first line, which means the number of test cases, and then n 32-bit integers follow.

Output

For each test case, you should output its reverse number, one case per line.

Sample Input

3

12

-12

1200

Sample Output

21

-21

2100

注意：前导0的情况！

例：

输入：

3

-0012560020

00000

00205

输出为：

-2006521

0

502

import java.util.Scanner;

public class Main{
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
while (t-- > 0) {
String str = sc.next();
int instr = Integer.parseInt(str);
//System.out.println(instr);
str = Integer.toString(instr);

//System.out.println(str);
if (str.charAt(0) == '-') {
System.out.print("-");
int k = 0;
boolean isOne=false;

//System.out.println(str.length()+"aaa");

for (int i = str.length() - 1; i >= 1; i--) {
//System.out.println("a:  "+str.charAt(i));
if(str.charAt(i)!='0'&&!isOne){
//System.out.println("++ "+str.charAt(i));
isOne=true;
}

if (isOne) {
System.out.print(str.charAt(i));
k++;
}
}
for (int i = 1; i < str.length() - k; i++) {
System.out.print(0);
}
System.out.println();
} else {
int k = 0;
boolean isOne=false;
for (int i = str.length() - 1; i >= 0; i--) {
if(str.charAt(i)!='0'&&!isOne){
isOne=true;
}

if (isOne) {
System.out.print(str.charAt(i));
k++;

}
}

for (int i = 0; i < str.length() - k; i++) {
System.out.print(0);
}
System.out.println();

}

}
}

}